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Exam questions · Maths · Graphs

Equations of Straight Lines

  • 6 exam questions
  • 21 marks
  • 9 quick checks
  1. 1 Work out [2 marks]

    \(A\) is the point \((2, 6)\) and \(B\) is the point \((8, 10)\). Find the coordinates of the midpoint of \(AB\).

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    Model answer

    \(\left(\dfrac{2 + 8}{2}, \dfrac{6 + 10}{2}\right) = (5, 8)\).

    Mark scheme

    • One coordinate correct — M1
    • \((5, 8)\) — A1
  2. 2 Work out [6 marks]

    The diagram shows a straight line through the points \(A\) and \(B\). (a) Work out the gradient of the line. (2 marks) (b) Find an equation of the line. (2 marks) (c) Work out the coordinates of the midpoint of \(AB\). (2 marks)

    A straight line passing through the points A (1, 3) and B (5, 11).
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    Model answer

    (a) \(\dfrac{11 - 3}{5 - 1} = 2\). (b) \(y = 2x + c\) with \((1, 3)\) gives \(3 = 2 + c\), so \(c = 1\) and \(y = 2x + 1\). (c) \(\left(\dfrac{1 + 5}{2}, \dfrac{3 + 11}{2}\right) = (3, 7)\).

    Mark scheme

    • (a) \(\dfrac{11 - 3}{5 - 1}\) — M1
    • (a) 2 — A1
    • (b) \(y = 2x + c\) and a point substituted — M1
    • (b) \(y = 2x + 1\) — A1
    • (c) One coordinate correct — M1
    • (c) \((3, 7)\) — A1
  3. 3 Find [3 marks]

    Find an equation of the line that is parallel to \(y = 3x - 1\) and passes through the point \((2, 9)\).

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    Model answer

    The gradient is 3, so \(y = 3x + c\). Substituting \((2, 9)\) gives \(9 = 6 + c\), so \(c = 3\) and \(y = 3x + 3\).

    Mark scheme

    • \(y = 3x + c\) or gradient 3 stated — M1
    • \(9 = 3 \times 2 + c\) — M1
    • \(y = 3x + 3\) — A1
  4. 4 Work out [3 marks]

    A line has equation \(3x + 2y = 12\). (a) Work out the gradient of the line. (2 marks) (b) Does the point \((2, 3)\) lie on the line? You must show your working. (1 mark)

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    Model answer

    (a) Rearrange to \(2y = -3x + 12\), so \(y = -\dfrac{3}{2}x + 6\). The gradient is \(-\dfrac{3}{2}\). (b) \(3 \times 2 + 2 \times 3 = 12\), so yes.

    Mark scheme

    • (a) \(2y = -3x + 12\) or \(y = \ldots\) — M1
    • (a) \(-\dfrac{3}{2}\) — A1
    • (b) Yes, with \(6 + 6 = 12\) — B1
  5. 5 Find [3 marks]

    Line \(L_1\) has equation \(y = 2x + 3\). Line \(L_2\) is perpendicular to \(L_1\) and passes through the point \((4, 1)\). Find an equation of \(L_2\).

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    Model answer

    The gradient of \(L_2\) is \(-\dfrac{1}{2}\). Then \(1 = -\dfrac{1}{2} \times 4 + c\), so \(c = 3\) and \(y = -\dfrac{1}{2}x + 3\).

    Mark scheme

    • Gradient \(-\dfrac{1}{2}\) seen — M1
    • \(1 = -\dfrac{1}{2} \times 4 + c\) — M1
    • \(y = -\dfrac{1}{2}x + 3\) — A1
  6. 6 Work out [4 marks]

    \(A\) is the point \((-2, 1)\) and \(B\) is the point \((6, 7)\). (a) Find the coordinates of the midpoint of \(AB\). (2 marks) (b) Work out the length of \(AB\). (2 marks)

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    Model answer

    (a) \(\left(\dfrac{-2 + 6}{2}, \dfrac{1 + 7}{2}\right) = (2, 4)\). (b) The differences are 8 and 6, so \(AB = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\).

    Mark scheme

    • (a) One coordinate correct — M1
    • (a) \((2, 4)\) — A1
    • (b) \(\sqrt{8^2 + 6^2}\) — M1
    • (b) 10 — A1

Quick check

  1. 1

    What is the equation of the line through \((2, 1)\) and \((6, 9)\)?

    1. A\(y = 2x + 3\)
    2. B\(y = \dfrac{1}{2}x - 3\)
    3. C\(y = 2x - 3\)
    4. D\(y = -2x - 3\)
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    C: \(y = 2x - 3\)

    The gradient is \(\dfrac{8}{4} = 2\). Then \(1 = 2 \times 2 + c\) gives \(c = -3\).

  2. 2

    What is the midpoint of \((2, 1)\) and \((6, 9)\)?

    1. A\((8, 10)\)
    2. B\((4, 5)\)
    3. C\((2, 4)\)
    4. D\((4, 8)\)
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    B: \((4, 5)\)

    \(\left(\dfrac{2 + 6}{2}, \dfrac{1 + 9}{2}\right) = (4, 5)\).

  3. 3

    A line is parallel to \(y = 5x - 2\). What is its gradient?

    1. A\(5\)
    2. B\(-5\)
    3. C\(-\dfrac{1}{5}\)
    4. D\(\dfrac{1}{5}\)
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    A: \(5\)

    Parallel lines have equal gradients.

  4. 4

    A line has gradient 3 and passes through \((2, 9)\). What is its equation?

    1. A\(y = 3x + 9\)
    2. B\(y = 3x - 3\)
    3. C\(y = 3x + 6\)
    4. D\(y = 3x + 3\)
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    D: \(y = 3x + 3\)

    \(9 = 3 \times 2 + c\) gives \(c = 3\).

  5. 5

    What is the gradient of the line \(2y - 4x = 6\)?

    1. A\(4\)
    2. B\(3\)
    3. C\(2\)
    4. D\(-2\)
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    C: \(2\)

    Rearranged, \(2y = 4x + 6\), so \(y = 2x + 3\).

  6. 6

    What is the midpoint of \((0, 4)\) and \((6, 10)\)?

    1. A\((6, 14)\)
    2. B\((3, 7)\)
    3. C\((3, 3)\)
    4. D\((6, 7)\)
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    B: \((3, 7)\)

    \(\left(\dfrac{0 + 6}{2}, \dfrac{4 + 10}{2}\right) = (3, 7)\).

  7. 7

    Where does the line \(3x + 2y = 12\) cross the axes?

    1. A\((0, 6)\) and \((4, 0)\)
    2. B\((0, 4)\) and \((6, 0)\)
    3. C\((0, 12)\) and \((12, 0)\)
    4. D\((0, 3)\) and \((2, 0)\)
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    A: \((0, 6)\) and \((4, 0)\)

    Put \(x = 0\) to get \(y = 6\), and \(y = 0\) to get \(x = 4\).

  8. 8

    What is the gradient of a line perpendicular to a line with gradient 4?

    1. A\(\dfrac{1}{4}\)
    2. B\(-4\)
    3. C\(4\)
    4. D\(-\dfrac{1}{4}\)
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    D: \(-\dfrac{1}{4}\)

    Perpendicular gradients multiply to \(-1\), so the new gradient is \(-\dfrac{1}{4}\).

  9. 9

    What is the equation of the line perpendicular to \(y = 2x + 3\) through \((4, 1)\)?

    1. A\(y = -2x + 9\)
    2. B\(y = \dfrac{1}{2}x - 1\)
    3. C\(y = -\dfrac{1}{2}x + 3\)
    4. D\(y = -\dfrac{1}{2}x + 1\)
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    C: \(y = -\dfrac{1}{2}x + 3\)

    The gradient is \(-\dfrac{1}{2}\). Then \(1 = -2 + c\), so \(c = 3\).